What can $\operatorname{Hom}\left(\prod_i M_i, N\right)$ look like? It's easy to see that $\operatorname{Hom}\left(\bigoplus_i M_i, N\right) = \prod_i \operatorname{Hom}(M_i, N)$.  However, there are a couple of ways this can conceivably fail if we replace the coproduct on the left with a product: we could have a homomorphism $\bigoplus_i M_i \rightarrow N$ which didn't extend to $\prod_i M_i$, or we could have one that extended non-uniquely.
In the latter case, there would be a nonzero homomorphism $\prod_i M_i \rightarrow N$ which was identically zero on elements of "finite support."  Can this happen, and if so is there a nice example?  (Apologies if this is obvious -- I've been thinking about it for a while and can't come up with anything.)
 A: By applying the hom functor to the exact sequence
$$ 0 \to \coprod_i M_i \to \prod_i M_i \to C \to 0 $$
(where $C$ is defined as the cokernel), we get a long exact sequence
$$ 0 \to \mathop{\mathrm{Hom}}(C, N) \to \mathop{\mathrm{Hom}}\left(\prod_i M_i, N\right) \to \mathop{\mathrm{Hom}}\left(\coprod_i M_i, N\right)
\\ \to \mathop{\mathrm{Ext^1}}(C, N) \to \mathop{\mathrm{Ext^1}}\left(\prod_i M_i, N\right) \to \mathop{\mathrm{Ext^1}}\left(\coprod_i M_i, N\right) \to \cdots $$
(the next term in the sequence is 0 if you're working with Abelian groups)
If you can understand the groups in this sequence, you can compute things. For example, understanding $C$ (and homomorphisms from $C$) tells you what homomorphisms $\prod_i M_i \to N$ vanish on elements of finite support. Or if $N$ is an injective module, then $\mathop{\mathrm{Ext^1}}(C, N) \cong 0$, and so every homomorphism $\coprod_i M_i \to N$ extends (probably not uniquely) to a homomorphism $\prod_i M_i \to N$. (this last statement, I believe, is not an 'if and only if')
(p.s. I really, really prefer limiting $\oplus$ to biproducts, and thus I don't like using it for infinite coproducts, which is why I use the notation $\coprod$ instead)
A: Here is a slightly simpler answer to your specific question. If your $M_i$ are modules over a ring, there is a canonical example of a homomorphism that kills the elements of finite support (which also works for many other algebraic structures): let $F \subseteq \prod_i M_i$ be the set of elements of finite support, then $F$ is a submodule; if you take $N$ to be the quotient module $(\prod_i M_i)/F$, then the projection $ p : \prod_i M_i \rightarrow N$  is identically zero on $F$. Any homomorphism on $\prod_i M_i$ that maps $F$ to zero factors through $p$. It may help to think of $N$ as what you get from the product by identifying elements that are "almost identical", where  $x$ and $y$ are "almost identical" if $x_i \not= y_i$ for only finitely many $i$.
A: I'm not sure whether nonzero homomorphisms $\prod_i M_i\to N$ can exist which are trivial on elements of finite support, but there are certainly homomorphisms $\bigoplus_i M_i\to N$ which do not extend to $\prod_i M_i\to N$. Namely, the homomorphism $\bigoplus_{i=1}^\infty M\to M$ given by $(m_1,m_2,\ldots)\mapsto \sum\limits_{i=1}^\infty m_i$ does not extend to $\prod_{i=1}^\infty M$.
